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Thanks for the help. It's behaving correctly now :)
1. N (bar number) and Price (Close[N])
The second line has coordinates
2. N-per(seed) and Price2(Close[N-per])
To get the third point as a continuation of this line, you must
to know the coordinates of the third point, which are
3. N-2*per and Price2(Close[N-per]) +DeltaPrice
DeltaPrice=Price2(Close[N-per])-Price (Close[N])
This is theory, the practice will be slightly different.
The second line has the coordinates
2. N-per(seed) and Price2(Close[N-per])
To get the third point as a continuation of this line, you must
to know the coordinates of the third point, which are
3. N-2*per and Price2(Close[N-per]) +DeltaPrice
DeltaPrice=Price2(Close[N-per])-Price (Close[N])
This is theory, the practice will be slightly different.
Good academic explanation :) Could you add a shift by N candles to the right in the same code. Of course, maybe it's not very modest of me, but I hope you will find the idea interesting :)
The easiest way to shift for a single buffer indicator is to add a second buffer in which you assign values to buffer2[i+N]=buffer1[i].
In doing so, suppress the output of the first buffer on the chart, and output the second buffer.
At the same time, suppress the output of the first buffer on the chart, and output the second one.
It is not just a shift, it is a 180-degree reverse of the first line itself. The picture shows the same first line, but rotated 180 degrees and highlighted in yellow.
Interesting link. I would like to discuss with you some issues of mathematical logic in an intimate atmosphere :) Which of your mailboxes can I use? You have our mailbox.